Question

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PROPER SOLUTION STEP BY STEP..........

ψ Q - 24 ψ

Answer 1

Qn:    2 (Sin⁶θ + Cos⁶θ) - 3 (Sin⁴ θ+ cos⁴θ) + 1 = 0

We know Sin² θ+ Cos²θ = 1
Squaring it on both sides, we get:
      Sin⁴ θ+ Cos⁴ θ = 1² -  2 Sin²θ Cos² θ
Cubing it on both sides, we get:
      Sin⁶θ + Cos⁶θ = 1³ - 3 Sin²θ Cos²θ (Sin²θ + Cos²θ) 
                              = 1 - 3 Sin²θ Cos²θ

Substituting it in the given equation:
LHS = 2 (1 - 3 Sin²θ Cos²θ) - 3 (1 - 2 Sin² θ Cos²θ) + 1 
        = 0          Proved

Answer 2

2(sin6x + cos6x) - 3(sin4x + cos4x)+1

=2[(sin2x)3+ (cos2x)3] - 3(sin4x + cos4x)+1

= 2(sin2x + cos2x)(sin4x - sin2xcos2x + cos4x)-3(sin4x + cos4x)+1 [a3+b3=(a+b)(a2-ab+b2)]

= 2(sin4x - sin2xcos2x + cos4x) - 3(sin4x + cos4x)+1 [As sin2x + cos2x=1]

= 2(sin4x + cos4x) - 2sin2xcos2x - 3(sin4x + cos4x)+1

= -(sin4x + cos4x) - 2sin2xcos2x+1

= -(sin4x + cos4x + 2sin2xcos2x)+1

= -[(sin2x)2+ (cos2x)2+ 2(sin2x)(cos2x)]+1

= -(sin2x + cos2x)2+1 [As a2+b2+2ab=(a+b)2]

= -(1)2+1 [As sin2x + cos2x=1]

= -1+1 =0

Hence, 2(sin6x + cos6x) - 3(sin4x + cos4x)+1=0

Hope its help you dear